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sin(4x-3)cos(x+5)
Solving integrals/ sin(4x-3)cos(x+5)

Integral of sin(4x-3)cos(x+5) dx

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∫01sin⁡(4x−3)cos⁡(x+5) dx\int\limits_{0}^{1} \sin{\left(4 x - 3 \right)} \cos{\left(x + 5 \right)}\, dx
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Detail solution

  1. Rewrite the integrand:

    sin⁡(4x−3)cos⁡(x+5)=8sin⁡(5)sin⁡4(x)cos⁡(3)cos⁡(x)−8sin⁡3(x)cos⁡(3)cos⁡(5)cos⁡2(x)−4sin⁡(5)sin⁡2(x)cos⁡(3)cos⁡(x)+8sin⁡(3)sin⁡(5)sin⁡(x)cos⁡4(x)+4sin⁡(x)cos⁡(3)cos⁡(5)cos⁡2(x)−8sin⁡(3)sin⁡(5)sin⁡(x)cos⁡2(x)+sin⁡(3)sin⁡(5)sin⁡(x)−8sin⁡(3)cos⁡(5)cos⁡5(x)+8sin⁡(3)cos⁡(5)cos⁡3(x)−sin⁡(3)cos⁡(5)cos⁡(x)\sin{\left(4 x - 3 \right)} \cos{\left(x + 5 \right)} = 8 \sin{\left(5 \right)} \sin^{4}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(x \right)} - 8 \sin^{3}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{2}{\left(x \right)} - 4 \sin{\left(5 \right)} \sin^{2}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(x \right)} + 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)} \cos^{4}{\left(x \right)} + 4 \sin{\left(x \right)} \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{2}{\left(x \right)} - 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)} - 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \cos^{5}{\left(x \right)} + 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \cos^{3}{\left(x \right)} - \sin{\left(3 \right)} \cos{\left(5 \right)} \cos{\left(x \right)}

  2. Integrate term-by-term:

    1. The integral of a constant times a function is the constant times the integral of the function:

      ∫8sin⁡(5)sin⁡4(x)cos⁡(3)cos⁡(x) dx=8sin⁡(5)cos⁡(3)∫sin⁡4(x)cos⁡(x) dx\int 8 \sin{\left(5 \right)} \sin^{4}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(x \right)}\, dx = 8 \sin{\left(5 \right)} \cos{\left(3 \right)} \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx

      1. Let u=sin⁡(x)u = \sin{\left(x \right)}.

        Then let du=cos⁡(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

        ∫u4 du\int u^{4}\, du

        1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

          ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

        Now substitute uu back in:

        sin⁡5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

      So, the result is: 8sin⁡(5)sin⁡5(x)cos⁡(3)5\frac{8 \sin{\left(5 \right)} \sin^{5}{\left(x \right)} \cos{\left(3 \right)}}{5}

    1. The integral of a constant times a function is the constant times the integral of the function:

      ∫(−8sin⁡3(x)cos⁡(3)cos⁡(5)cos⁡2(x)) dx=−8cos⁡(3)cos⁡(5)∫sin⁡3(x)cos⁡2(x) dx\int \left(- 8 \sin^{3}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 8 \cos{\left(3 \right)} \cos{\left(5 \right)} \int \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

      1. Rewrite the integrand:

        sin⁡3(x)cos⁡2(x)=(1−cos⁡2(x))sin⁡(x)cos⁡2(x)\sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)}

      2. There are multiple ways to do this integral.

        Method #1

        1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

          Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute dudu:

          ∫(u4−u2) du\int \left(u^{4} - u^{2}\right)\, du

          1. Integrate term-by-term:

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

              ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

            1. The integral of a constant times a function is the constant times the integral of the function:

              ∫(−u2) du=−∫u2 du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

              1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

              So, the result is: −u33- \frac{u^{3}}{3}

            The result is: u55−u33\frac{u^{5}}{5} - \frac{u^{3}}{3}

          Now substitute uu back in:

          cos⁡5(x)5−cos⁡3(x)3\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}

        Method #2

        1. Rewrite the integrand:

          (1−cos⁡2(x))sin⁡(x)cos⁡2(x)=−sin⁡(x)cos⁡4(x)+sin⁡(x)cos⁡2(x)\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)} = - \sin{\left(x \right)} \cos^{4}{\left(x \right)} + \sin{\left(x \right)} \cos^{2}{\left(x \right)}

        2. Integrate term-by-term:

          1. The integral of a constant times a function is the constant times the integral of the function:

            ∫(−sin⁡(x)cos⁡4(x)) dx=−∫sin⁡(x)cos⁡4(x) dx\int \left(- \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx

            1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

              Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

              ∫u4 du\int u^{4}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                ∫(−u4) du=−∫u4 du\int \left(- u^{4}\right)\, du = - \int u^{4}\, du

                1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                  ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

                So, the result is: −u55- \frac{u^{5}}{5}

              Now substitute uu back in:

              −cos⁡5(x)5- \frac{\cos^{5}{\left(x \right)}}{5}

            So, the result is: cos⁡5(x)5\frac{\cos^{5}{\left(x \right)}}{5}

          1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

            Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

            ∫u2 du\int u^{2}\, du

            1. The integral of a constant times a function is the constant times the integral of the function:

              ∫(−u2) du=−∫u2 du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

              1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

              So, the result is: −u33- \frac{u^{3}}{3}

            Now substitute uu back in:

            −cos⁡3(x)3- \frac{\cos^{3}{\left(x \right)}}{3}

          The result is: cos⁡5(x)5−cos⁡3(x)3\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}

        Method #3

        1. Rewrite the integrand:

          (1−cos⁡2(x))sin⁡(x)cos⁡2(x)=−sin⁡(x)cos⁡4(x)+sin⁡(x)cos⁡2(x)\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)} = - \sin{\left(x \right)} \cos^{4}{\left(x \right)} + \sin{\left(x \right)} \cos^{2}{\left(x \right)}

        2. Integrate term-by-term:

          1. The integral of a constant times a function is the constant times the integral of the function:

            ∫(−sin⁡(x)cos⁡4(x)) dx=−∫sin⁡(x)cos⁡4(x) dx\int \left(- \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx

            1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

              Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

              ∫u4 du\int u^{4}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                ∫(−u4) du=−∫u4 du\int \left(- u^{4}\right)\, du = - \int u^{4}\, du

                1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                  ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

                So, the result is: −u55- \frac{u^{5}}{5}

              Now substitute uu back in:

              −cos⁡5(x)5- \frac{\cos^{5}{\left(x \right)}}{5}

            So, the result is: cos⁡5(x)5\frac{\cos^{5}{\left(x \right)}}{5}

          1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

            Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

            ∫u2 du\int u^{2}\, du

            1. The integral of a constant times a function is the constant times the integral of the function:

              ∫(−u2) du=−∫u2 du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

              1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

              So, the result is: −u33- \frac{u^{3}}{3}

            Now substitute uu back in:

            −cos⁡3(x)3- \frac{\cos^{3}{\left(x \right)}}{3}

          The result is: cos⁡5(x)5−cos⁡3(x)3\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}

      So, the result is: −8(cos⁡5(x)5−cos⁡3(x)3)cos⁡(3)cos⁡(5)- 8 \left(\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}\right) \cos{\left(3 \right)} \cos{\left(5 \right)}

    1. The integral of a constant times a function is the constant times the integral of the function:

      ∫(−4sin⁡(5)sin⁡2(x)cos⁡(3)cos⁡(x)) dx=−4sin⁡(5)cos⁡(3)∫sin⁡2(x)cos⁡(x) dx\int \left(- 4 \sin{\left(5 \right)} \sin^{2}{\left(x \right)} \cos{\left(3 \right)} \cos{\left(x \right)}\right)\, dx = - 4 \sin{\left(5 \right)} \cos{\left(3 \right)} \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx

      1. Let u=sin⁡(x)u = \sin{\left(x \right)}.

        Then let du=cos⁡(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

        ∫u2 du\int u^{2}\, du

        1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

          ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

        Now substitute uu back in:

        sin⁡3(x)3\frac{\sin^{3}{\left(x \right)}}{3}

      So, the result is: −4sin⁡(5)sin⁡3(x)cos⁡(3)3- \frac{4 \sin{\left(5 \right)} \sin^{3}{\left(x \right)} \cos{\left(3 \right)}}{3}

    1. The integral of a constant times a function is the constant times the integral of the function:

      ∫8sin⁡(3)sin⁡(5)sin⁡(x)cos⁡4(x) dx=8sin⁡(3)sin⁡(5)∫sin⁡(x)cos⁡4(x) dx\int 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx

      1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

        Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

        ∫u4 du\int u^{4}\, du

        1. The integral of a constant times a function is the constant times the integral of the function:

          ∫(−u4) du=−∫u4 du\int \left(- u^{4}\right)\, du = - \int u^{4}\, du

          1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

            ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

          So, the result is: −u55- \frac{u^{5}}{5}

        Now substitute uu back in:

        −cos⁡5(x)5- \frac{\cos^{5}{\left(x \right)}}{5}

      So, the result is: −8sin⁡(3)sin⁡(5)cos⁡5(x)5- \frac{8 \sin{\left(3 \right)} \sin{\left(5 \right)} \cos^{5}{\left(x \right)}}{5}

    1. The integral of a constant times a function is the constant times the integral of the function:

      ∫4sin⁡(x)cos⁡(3)cos⁡(5)cos⁡2(x) dx=4cos⁡(3)cos⁡(5)∫sin⁡(x)cos⁡2(x) dx\int 4 \sin{\left(x \right)} \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{2}{\left(x \right)}\, dx = 4 \cos{\left(3 \right)} \cos{\left(5 \right)} \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

      1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

        Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

        ∫u2 du\int u^{2}\, du

        1. The integral of a constant times a function is the constant times the integral of the function:

          ∫(−u2) du=−∫u2 du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

          1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

            ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

          So, the result is: −u33- \frac{u^{3}}{3}

        Now substitute uu back in:

        −cos⁡3(x)3- \frac{\cos^{3}{\left(x \right)}}{3}

      So, the result is: −4cos⁡(3)cos⁡(5)cos⁡3(x)3- \frac{4 \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{3}{\left(x \right)}}{3}

    1. The integral of a constant times a function is the constant times the integral of the function:

      ∫(−8sin⁡(3)sin⁡(5)sin⁡(x)cos⁡2(x)) dx=−8sin⁡(3)sin⁡(5)∫sin⁡(x)cos⁡2(x) dx\int \left(- 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 8 \sin{\left(3 \right)} \sin{\left(5 \right)} \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

      1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

        Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

        ∫u2 du\int u^{2}\, du

        1. The integral of a constant times a function is the constant times the integral of the function:

          ∫(−u2) du=−∫u2 du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

          1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

            ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

          So, the result is: −u33- \frac{u^{3}}{3}

        Now substitute uu back in:

        −cos⁡3(x)3- \frac{\cos^{3}{\left(x \right)}}{3}

      So, the result is: 8sin⁡(3)sin⁡(5)cos⁡3(x)3\frac{8 \sin{\left(3 \right)} \sin{\left(5 \right)} \cos^{3}{\left(x \right)}}{3}

    1. The integral of a constant times a function is the constant times the integral of the function:

      ∫sin⁡(3)sin⁡(5)sin⁡(x) dx=sin⁡(3)sin⁡(5)∫sin⁡(x) dx\int \sin{\left(3 \right)} \sin{\left(5 \right)} \sin{\left(x \right)}\, dx = \sin{\left(3 \right)} \sin{\left(5 \right)} \int \sin{\left(x \right)}\, dx

      1. The integral of sine is negative cosine:

        ∫sin⁡(x) dx=−cos⁡(x)\int \sin{\left(x \right)}\, dx = - \cos{\left(x \right)}

      So, the result is: −sin⁡(3)sin⁡(5)cos⁡(x)- \sin{\left(3 \right)} \sin{\left(5 \right)} \cos{\left(x \right)}

    1. The integral of a constant times a function is the constant times the integral of the function:

      ∫(−8sin⁡(3)cos⁡(5)cos⁡5(x)) dx=−8sin⁡(3)cos⁡(5)∫cos⁡5(x) dx\int \left(- 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \cos^{5}{\left(x \right)}\right)\, dx = - 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \int \cos^{5}{\left(x \right)}\, dx

      1. Rewrite the integrand:

        cos⁡5(x)=(1−sin⁡2(x))2cos⁡(x)\cos^{5}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)}

      2. Rewrite the integrand:

        (1−sin⁡2(x))2cos⁡(x)=sin⁡4(x)cos⁡(x)−2sin⁡2(x)cos⁡(x)+cos⁡(x)\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)} = \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}

      3. Integrate term-by-term:

        1. Let u=sin⁡(x)u = \sin{\left(x \right)}.

          Then let du=cos⁡(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

          ∫u4 du\int u^{4}\, du

          1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

            ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

          Now substitute uu back in:

          sin⁡5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

        1. The integral of a constant times a function is the constant times the integral of the function:

          ∫(−2sin⁡2(x)cos⁡(x)) dx=−2∫sin⁡2(x)cos⁡(x) dx\int \left(- 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 2 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx

          1. Let u=sin⁡(x)u = \sin{\left(x \right)}.

            Then let du=cos⁡(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

            ∫u2 du\int u^{2}\, du

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

              ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

            Now substitute uu back in:

            sin⁡3(x)3\frac{\sin^{3}{\left(x \right)}}{3}

          So, the result is: −2sin⁡3(x)3- \frac{2 \sin^{3}{\left(x \right)}}{3}

        1. The integral of cosine is sine:

          ∫cos⁡(x) dx=sin⁡(x)\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}

        The result is: sin⁡5(x)5−2sin⁡3(x)3+sin⁡(x)\frac{\sin^{5}{\left(x \right)}}{5} - \frac{2 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}

      So, the result is: −8(sin⁡5(x)5−2sin⁡3(x)3+sin⁡(x))sin⁡(3)cos⁡(5)- 8 \left(\frac{\sin^{5}{\left(x \right)}}{5} - \frac{2 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}\right) \sin{\left(3 \right)} \cos{\left(5 \right)}

    1. The integral of a constant times a function is the constant times the integral of the function:

      ∫8sin⁡(3)cos⁡(5)cos⁡3(x) dx=8sin⁡(3)cos⁡(5)∫cos⁡3(x) dx\int 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \cos^{3}{\left(x \right)}\, dx = 8 \sin{\left(3 \right)} \cos{\left(5 \right)} \int \cos^{3}{\left(x \right)}\, dx

      1. Rewrite the integrand:

        cos⁡3(x)=(1−sin⁡2(x))cos⁡(x)\cos^{3}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}

      2. Let u=sin⁡(x)u = \sin{\left(x \right)}.

        Then let du=cos⁡(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

        ∫(1−u2) du\int \left(1 - u^{2}\right)\, du

        1. Integrate term-by-term:

          1. The integral of a constant is the constant times the variable of integration:

            ∫1 du=u\int 1\, du = u

          1. The integral of a constant times a function is the constant times the integral of the function:

            ∫(−u2) du=−∫u2 du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

              ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

            So, the result is: −u33- \frac{u^{3}}{3}

          The result is: −u33+u- \frac{u^{3}}{3} + u

        Now substitute uu back in:

        −sin⁡3(x)3+sin⁡(x)- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}

      So, the result is: 8(−sin⁡3(x)3+sin⁡(x))sin⁡(3)cos⁡(5)8 \left(- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}\right) \sin{\left(3 \right)} \cos{\left(5 \right)}

    1. The integral of a constant times a function is the constant times the integral of the function:

      ∫(−sin⁡(3)cos⁡(5)cos⁡(x)) dx=−sin⁡(3)cos⁡(5)∫cos⁡(x) dx\int \left(- \sin{\left(3 \right)} \cos{\left(5 \right)} \cos{\left(x \right)}\right)\, dx = - \sin{\left(3 \right)} \cos{\left(5 \right)} \int \cos{\left(x \right)}\, dx

      1. The integral of cosine is sine:

        ∫cos⁡(x) dx=sin⁡(x)\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}

      So, the result is: −sin⁡(3)sin⁡(x)cos⁡(5)- \sin{\left(3 \right)} \sin{\left(x \right)} \cos{\left(5 \right)}

    The result is: 8(−sin⁡3(x)3+sin⁡(x))sin⁡(3)cos⁡(5)−8(cos⁡5(x)5−cos⁡3(x)3)cos⁡(3)cos⁡(5)−8(sin⁡5(x)5−2sin⁡3(x)3+sin⁡(x))sin⁡(3)cos⁡(5)+8sin⁡(5)sin⁡5(x)cos⁡(3)5−4sin⁡(5)sin⁡3(x)cos⁡(3)3−sin⁡(3)sin⁡(x)cos⁡(5)−8sin⁡(3)sin⁡(5)cos⁡5(x)5+8sin⁡(3)sin⁡(5)cos⁡3(x)3−4cos⁡(3)cos⁡(5)cos⁡3(x)3−sin⁡(3)sin⁡(5)cos⁡(x)8 \left(- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}\right) \sin{\left(3 \right)} \cos{\left(5 \right)} - 8 \left(\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}\right) \cos{\left(3 \right)} \cos{\left(5 \right)} - 8 \left(\frac{\sin^{5}{\left(x \right)}}{5} - \frac{2 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}\right) \sin{\left(3 \right)} \cos{\left(5 \right)} + \frac{8 \sin{\left(5 \right)} \sin^{5}{\left(x \right)} \cos{\left(3 \right)}}{5} - \frac{4 \sin{\left(5 \right)} \sin^{3}{\left(x \right)} \cos{\left(3 \right)}}{3} - \sin{\left(3 \right)} \sin{\left(x \right)} \cos{\left(5 \right)} - \frac{8 \sin{\left(3 \right)} \sin{\left(5 \right)} \cos^{5}{\left(x \right)}}{5} + \frac{8 \sin{\left(3 \right)} \sin{\left(5 \right)} \cos^{3}{\left(x \right)}}{3} - \frac{4 \cos{\left(3 \right)} \cos{\left(5 \right)} \cos^{3}{\left(x \right)}}{3} - \sin{\left(3 \right)} \sin{\left(5 \right)} \cos{\left(x \right)}

  3. Now simplify:

    −2cos⁡2(2x)cos⁡(x+2)5−4cos⁡(2x)cos⁡(x−2)5+3cos⁡(x+2)5−cos⁡(3x−8)6+cos⁡(3x−2)2- \frac{2 \cos^{2}{\left(2 x \right)} \cos{\left(x + 2 \right)}}{5} - \frac{4 \cos{\left(2 x \right)} \cos{\left(x - 2 \right)}}{5} + \frac{3 \cos{\left(x + 2 \right)}}{5} - \frac{\cos{\left(3 x - 8 \right)}}{6} + \frac{\cos{\left(3 x - 2 \right)}}{2}

  4. Add the constant of integration:

    −2cos⁡2(2x)cos⁡(x+2)5−4cos⁡(2x)cos⁡(x−2)5+3cos⁡(x+2)5−cos⁡(3x−8)6+cos⁡(3x−2)2+constant- \frac{2 \cos^{2}{\left(2 x \right)} \cos{\left(x + 2 \right)}}{5} - \frac{4 \cos{\left(2 x \right)} \cos{\left(x - 2 \right)}}{5} + \frac{3 \cos{\left(x + 2 \right)}}{5} - \frac{\cos{\left(3 x - 8 \right)}}{6} + \frac{\cos{\left(3 x - 2 \right)}}{2}+ \mathrm{constant}

The answer is:

−2cos⁡2(2x)cos⁡(x+2)5−4cos⁡(2x)cos⁡(x−2)5+3cos⁡(x+2)5−cos⁡(3x−8)6+cos⁡(3x−2)2+constant- \frac{2 \cos^{2}{\left(2 x \right)} \cos{\left(x + 2 \right)}}{5} - \frac{4 \cos{\left(2 x \right)} \cos{\left(x - 2 \right)}}{5} + \frac{3 \cos{\left(x + 2 \right)}}{5} - \frac{\cos{\left(3 x - 8 \right)}}{6} + \frac{\cos{\left(3 x - 2 \right)}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                 /     3         5   \                   /       3         5            \                   /     3            \                      5                         3                         3                         3                         5                 
 |                                                                                  |  cos (x)   cos (x)|                   |  2*sin (x)   sin (x)         |                   |  sin (x)         |                 8*cos (x)*sin(3)*sin(5)   4*cos (x)*cos(3)*cos(5)   4*sin (x)*cos(3)*sin(5)   8*cos (x)*sin(3)*sin(5)   8*sin (x)*cos(3)*sin(5)
 | sin(4*x - 3)*cos(x + 5) dx = C - cos(5)*sin(3)*sin(x) - cos(x)*sin(3)*sin(5) - 8*|- ------- + -------|*cos(3)*cos(5) - 8*|- --------- + ------- + sin(x)|*cos(5)*sin(3) + 8*|- ------- + sin(x)|*cos(5)*sin(3) - ----------------------- - ----------------------- - ----------------------- + ----------------------- + -----------------------
 |                                                                                  \     3         5   /                   \      3          5            /                   \     3            /                            5                         3                         3                         3                         5           
/
−cos⁡(5 x+2)10−cos⁡(3 x−8)6-{{\cos \left(5\,x+2\right)}\over{10}}-{{\cos \left(3\,x-8\right) }\over{6}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.902-2
Plot the graph f(x)
The answer [src] 
  4*cos(1)*cos(6)   sin(1)*sin(6)   sin(3)*sin(5)   4*cos(3)*cos(5)
- --------------- - ------------- - ------------- + ---------------
         15               15              15               15
5 cos⁡8+3 cos⁡230−3 cos⁡7+5 cos⁡530{{5\,\cos 8+3\,\cos 2}\over{30}}-{{3\,\cos 7+5\,\cos 5}\over{30}} 
=
= 
  4*cos(1)*cos(6)   sin(1)*sin(6)   sin(3)*sin(5)   4*cos(3)*cos(5)
- --------------- - ------------- - ------------- + ---------------
         15               15              15               15
−4cos⁡(1)cos⁡(6)15+4cos⁡(3)cos⁡(5)15−sin⁡(3)sin⁡(5)15−sin⁡(1)sin⁡(6)15- \frac{4 \cos{\left(1 \right)} \cos{\left(6 \right)}}{15} + \frac{4 \cos{\left(3 \right)} \cos{\left(5 \right)}}{15} - \frac{\sin{\left(3 \right)} \sin{\left(5 \right)}}{15} - \frac{\sin{\left(1 \right)} \sin{\left(6 \right)}}{15}
Simplify
Numerical answer [src] 
-0.188531945634351
-0.188531945634351
The graph
Integral of sin(4x-3)cos(x+5) dx

    Use the examples entering the upper and lower limits of integration.

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